GATE | Gate IT 2007 | Question 11
Let a memory have four free blocks of sizes 4k, 8k, 20k, 2k. These blocks are allocated following the best-fit strategy. The allocation requests are stored in a queue as shown below.
The time at which the request for J7 will be completed will be
Explanation: Initially when a process arrives and needs memory, it would search for a hole big enough to fit the job and if the hole is larger then the remaining hole is returned to the free storage list.
|Memory Block||Size||Job (t=0)||Job(t=8)||Job(t=10)||Job(t=11)|
|1||4k||J3 – 2 units (1K free left)|
|2||8k||J4 – 8 units (2K free left)||J5 – 14 units||J5 – 14 units||J5 – 14 units|
|3||20k||J2 -10 units(6K free left)||J2 -10 units||J6 – 11 units||J7 – 19 units|
|4||2k||J1 -4 units|
Therefore, the process finishes at J7=19 units